Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.util.Locale;
import java.util.Scanner;

public class Main {
    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in).useLocale(Locale.ENGLISH);
        int count = scanner.nextInt();
        double[] arr = new double[count + 1];
        for (int i = 1; i <= count; ++i) {
            arr[i] = scanner.nextDouble();
        }

        double[][] dp = new double[count + 1][count + 1];
        double res = 0.0;
        dp[0][0] = 1.0;
        for (int i = 0; i <= count; ++i) {
            for (int j = 0; i + j <= count; ++j) {
                if (i > 0) {
                    dp[i][j] += dp[i - 1][j] * arr[i + j];
                }
                if (j > 0) {
                    dp[i][j] += dp[i][j - 1] * (1.0 - arr[i + j]);
                }
                if (i + j == count && i > j) {
                    res += dp[i][j];
                }
            }
        }

        System.out.println(res);
    }
}